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Total moment of inertia of a two particle system 
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#1
Oct2409, 04:49 PM

P: 190

1. Find the moment of inertia I_x of particle a with respect to the x axis (that is, if the x axis is the axis of rotation), the moment of inertia I_y of particle a with respect to the y axis, and the moment of inertia I_z of particle a with respect to the z axis (the axis that passes through the origin perpendicular to both the x and y axes).
Particle a is located 3r from the y axis, and particle b is located r away. 2. I = mr^2 I = SUMM(m_i*r_i^2) 3. I_a = 9mr^2 I_b = mr^2 Total I = 10mr^2 No idea what I am doing wrong 


#3
Oct2409, 04:57 PM

P: 190

Particle a:
distance from y axis = 3r distance from x axis = r Particle b: distance from y axis = r distance from x axis = 4r It doesn't say anything explicitly about mass, and so I think we assume each particle has mass m. 


#4
Oct2409, 05:02 PM

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Total moment of inertia of a two particle system



#5
Oct2409, 05:06 PM

P: 190

I_x would be m(r)^2 + m(4r)^2 = mr^2 + 16mr^2 = 17mr^2.
Am I simply adding that to I_y to get I_total? It asks for total I with respect to the y axis, though, why would I need I_x? 


#6
Oct2409, 05:10 PM

P: 190

Shoot, just realized I posted to wrong question. Here's what I am trying to answer:
Find the total moment of inertia I of the system of two particles shown in the diagram with respect to the y axis 


#7
Oct2409, 05:25 PM

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#8
Oct2409, 05:26 PM

P: 190




#10
Oct2409, 05:35 PM

P: 190

Ok, I requested the correct answer: 11mr^2
WTF?! 


#11
Oct2409, 05:37 PM

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#12
Oct2409, 05:39 PM

P: 190

Ok, now:
Using the formula for kinetic energy of a moving particle K=\frac{_1}{^2}mv^2, find the kinetic energy K_a of particle a and the kinetic energy K_b of particle b. Express your answers in terms of m, omega, and r separated by a comma. I know I need to calculate linear speed, how do I do that? I know it's v=omega*r but I don't see how to put this together..... 


#13
Oct2409, 05:42 PM

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#14
Oct2409, 05:44 PM

P: 190

Just edited before you replied: v=omega*r
But what are omega and r here?! 


#15
Oct2409, 05:49 PM

P: 190

Particle A:
Ka = 0.5m(3*omega*r)^2 Kb = 0.5m(omega*r)^2 Apparently Kb is wrong but I did it the same way as Ka............. 


#16
Oct2409, 05:58 PM

P: 190

Ok I got Kb = m(omega*r)^2
What is total kinteic eneregy? 


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